Can $0$ be an accumulation point in the spectrum of a finite-rank operator?

Question

Let $V$ be an infinite-dimensional Banach space and $T$ a finite-rank operator $V\to V$.

Question: Can $0$ be an accumulation point in the spectrum?

Thoughts: I understand that since $T$ is compact, $0$ must be in its spectrum, and for any $c>0$ the number of spectral points $z$ with $|z|>c$ is finite. So this is equivalent to asking whether the spectrum of $T$ is finite.

Answer 1

Let $M$ be the range of $T$ and $S$ be the restrictoon of $T$ to $M$. Then $S$ is an operator from $M$ into $M$ and any eigen value of $T$ other than $0$ is also an eigen value of $S$. [$Tx=\lambda x, \lambda \neq 0$ impies $x=\frac 1 {\lambda} Tx \in M$ and $Sx=\lambda x$]. Hence, there can only be finitely many non-zero eigen values. The maximum number of non-zero eigen vales is the dimension of $M$. Since $T$ is compact, the nozero eigen values are the only non-zero points in the spectrum of $T$.

Answer 2

Since $T$ is compact, the only non-zero elements of its spectrum are its non-zero eigenvalues. The direct sum of the corresponding eigen-subspaces being finite-dimensional (as a subspace of $\operatorname{im}T$), the spectrum is indeed finite.